Hi

I am using this common anode 7 segment display:
http://www.us.kingbright.com/images/catalog/spec/SA08-11EWA.pdf

I have it connected to this decoder:
http://www.farnell.com/datasheets/1700457.pdf

I eventually realised that the logic is good it's just that all the led's are inverted!

So a number 3 looks like a number 1, but aligned to the left!! The number 4 was just seemingly random leds, but I have realised that its just that the LEDs are inverted.

I said the same thing twice. Anyway...

Could anyone take a guess why all LEds are inverted?

I can probe the pins and all the logic seems sound, highs and lows where expected.

 

Look at page 1, lower right just above the chip outline. You are using LEDs with a chip designed to drive Liquid crystal displays. It will work but you need a transistor driver for each segment which will invert the output signal from the decoder. Look at the truth table for the various digits and this will become more clear.

 

Thanks would something like this work:
http://cpc.farnell.com/1/1/60374-darlington-array-8npn-2804-dip18-uln2804a-stmicroelectronics.html

So its not just a case of adding an inverter like this:
http://cpc.farnell.com/nxp/74hc04n/74hc-cmos-74hc04-dip14-5v/dp/SC08936

 

hi dux,
How is pin #6 , the PH, connected on the 4543 IC.?

E

 

I have it with a 1.5k resistor pulled down to ground

 

I have it with a 1.5k resistor pulled down to ground

Try the LED display when Pin 6 is pulled High.

Recheck the LED digits displayed.

 

Legend thanks

sorted

 

Legend thanks

sorted

Hi,
This image shows the result of connection to High of the phase.

You have used the internal XNOR to invert the output, the same as adding an external inverter.

E
EDIT: The Phase pin is normally switched by the Back Plane oscillator when driving an LCD, as the LCD segments require an 'AC' waveform drive.

 

So imagining that the XNOR is not present would I use an inverter :

http://cpc.farnell.com/nxp/74hc04n/74hc-cmos-74hc04-dip14-5v/dp/SC08936


or a darlington array :

http://cpc.farnell.com/1/1/60374-darlington-array-8npn-2804-dip18-uln2804a-stmicroelectronics.html

or either or neither

cheers

 

hi,
The HC04 can sink/source a max of 25mA on any one pin, but you would have to ensure that the max Wattage did not exceed the rating of the IC.

The ULN would easily sink the LED segment current, that would have been my choice.
E

 

Or change to a common cathode LED display instead?